How do you find an equation of hyperbola with given foci and asymptote: Foci: (0,0) and (0,4) Asymptote: y= +/-[(1/2)]x + 2?

1 Answer
Aug 9, 2017

The equation of the hyperbola is (y-2)^2-(x^2/4)=1

Explanation:

The foci are F=(0,4) and F'=(0,0)

The center is C=(0,2)

The equations of the asymptotes are

y=1/2x+2 and y=-1/2x+2

Therefore,

y-2=+-1/2x

Squaring both sides

(y-2)^2-(x^2/4)=0

Therefore,

The equation of the hyperbola is

(y-2)^2-(x^2/4)=1

Verification

The general equation of the hyperbola is

(y-h)^2/a^2-(x-k)^2/b^2=1

The foci are F=(k,h+c)=(0,2+2)=(0,4) and

F'=(k,h-c)=(0,2-2)=(0,0)

graph{((y-2)^2-(x^2)/4-1)(y-2-1/2x)(y-2+1/2x)=0 [-6.76, 7.28, -1.425, 5.6]}